the fourier series expansion of a real periodic signal with fundamental frequency f0 is given by

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26. The Laplace transform of i(t) is given by i(s) = 2/s(1 + s). as t the value of i(t) tends to

(a) 0

(b) 1

(c) 2

(d) 0

  1. The Fourier series expansion of a real periodic signal with fundamental frequency fo is given by gp(t) = c0ej2nft . it is given that c3 = 3 + j5.

(a) 5 + j3

(b) -3 – j5

(c) -5 + j3

(d) 3 – j5

  1. Let x(t) be the input to a linear, time-invariant system. the required output is 4x(t – 2). the transfer function of the system should be

(a) 4ej4t

(b) 2e-j8t

(c) 4e-j4t

(d) 2ej8t

  1. A sequence x(n) with the z-transform x(z) = z4 + z2 – 2z + 2 – 3z-4 is applied as an input to a linear, time-invariant sysyem with the impulse response h(n) = 2(n – 3), where (n) = {1, n = 0 0, otherwise

the output at n = 4 is

(a) -6

(b) zero

(c) 2

(d) -4

  1. Convolution of x(t + 5) with impulse function (t – 7) is equal to

(a) x(t – 12)

(b) x(t + 12)

(c) x(t – 2)

(d) x(t + 2)

  1. Which of the following cannot be the fourier series expansion of a periodic signal?

(a) x(t) = 2 cos t + 3 cos 3t

(b) x(t) = 2 cos t + 7 cos t

(c) x(t) = cos t + 0.5

(d) x(t) = 2 cos 1.5t + sin 3.5 t

  1. The fourier transform f(e-t u(t) is equal to 1/1 + j2f. therefore, f{1/1 – j2t} is

(a) ef u(f)

(b) e-f u(f)

(c) ef u(-f)

(d) e-f u(-f)

  1. A linear phase channel with phase delay tp and group delay tg must have

(a) tp = tg = constant

(b) tp f and tg f

(c) tp = constanf and tg f

(d) tp f and tg = constant

  1. Conider a sampled signal :

y(t) = 5 x 10-6 x(t)  (t – nts),

where,   x(t) = 10 cos (8 x 103) t

and ts = 100 us. when y(t) is passed through an ideal low-pass filter with a cut-off frequency of 5 khz, the output of the filter is

(a) 5 x 10-6 cos (8 x 103) t

(b) 5 x 10-5 cos (8 x 103) t

(c) 5 x 10-1 cos (8 x 103) t

(d) 10 cos (8 x 103) t

  1. The transfer function of a system is given by h(s) = 1/s2 (s – 2) . the impulse response of the system is (* denotes convolution, and u(t) is unit step function)

(a) (t2 * e-2t) u(t)

(b) (t * e2t) u(t)

(c) te-2t) u(t)

(d) (te-2t) u(t)

  1. The region of convergence of the z-transform of a unit step function is

(a) |z|>1

(b) |z|<1

(c) (real part of z)> 0

(d) (real part of z) < 0

  1. Let (t) denotes the delta function. the value of the integral (t) cos (3t/2) dt is

(a) 1

(b) -1

(c) 0

(d) 2

  1. A band limited signal is sampled at the nyquist rate. the signal can be recovered by passing the samples through

(a) an R-C filter

(b) an envelope detector

(c) a PLL

(d) an ideal low-pass filter with appropriate bandwidth

  1. A linear time-invariant system has an impulse response e2t, t > 0. if the initial conditions are zero and the input is e3t, the output for t > 0 is

(a) e3t – e2t

(b) e5t

(c) e3t + e3t

(d) none of these

  1. Given that l[f(t)] = s + 2/s2 + 1, l[g(t)] = s2 + 1/(s + 3) (s + 2). h(t) = f(r) g(t) d. l[h(t)] is

(a) s2 + 1/s + 3

(b) 1/s + 3

(c) s2 + 1/(s + 3) (s + 2) + s + 2/s2 + 1

(d) none of these

  1. The fourier transform of the signal x(t) = e-3t2 is of the following form, where A and B are constants

(a) Ae-5(t)

(b) Ae-bf2

(c) A + B |F|2

(d) Ae-bf

  1. A system with an input x(t) and output y(t) is described by the relation y(t) = tx(t). this system is

(a) linear and time-invariant

(b) linear and time-variant

(c) non-linear and time-invariant

(d) non-linear and time-variant

  1. The n-point discrete fourier transform of a signal x[n] is xdft [k]. if DFT of another discrete sequence y[n] is given as ydft [k] = x*dft [k], then which of the following is true?

(a) y[n] = x* [n]

(b) y[n] = nx* [n]

(c) y[n] = 1/n x* [n]

(d) y[n] = – x* [n]

  1. Discrete fourier transform of a real sequence {x[n]} is given as

xdft[k] = {0, a, 2 + j, – 1,b,j}

then, a and b are

(a) 2 – j, j

(b) j, -j

(c) -j, 2 – j

(d) j, 2 + j

  1. The 4-point discrete fourier transform of a discrete time sequence (1,2,1,2) is

(a) {0, -2, 0, 6}

(b) {6, -4j, 6 – 4j}

(c) {-4j , 6 – 4j, 6}

(d) {6, 0, -2, 0}

  1. Consider a band-pass signal xc (t) is shown below. the minimum sampling rate required for this signal to prevent loss of information is

(a) 4 kHz

(b) 12 kHz

(c) 20 kHz

(d) 2 kHz

  1. Two signals x1(t) are band-limited to 2 kHz and 3kHz respectively, then nyquist rate for the signal x1(t)* x2(t) is

(a) 5 kHz

(b) 6 kHz

(c) 4 kHz

(d) 10 kHz

  1. Which one is most appropriate dynamic system out of the following?

(a) y(n) = y(n – 1) + y(n + 1)

(b) y(n) = y(n – 1)

(c) y(n) = x(n)

(d) y(n) + y(n – 1) + y(n + 3) = 0

  1. Which one is causal system?

(a) y(n) = 3x(n) – 2x(n – 1)

(b) y(n) = 3x(n) + 2x(n + 1)

(c) y(n) = 3x(n + 1) + 2x(n – 1)

(d) y(n) = 3x(n + 1) + 2x(n – 1) + x (n)

  1. For the signal given below

y(t) = 2 sin (2/3 t) + 4sin (1/4 t 4) + 6 sin (1/3 t 5) + 8 sin (1/2 t 7)

the common period of y(t) is given by

(a) 12

(b) 24

(c) 8

(d) 16

 

  1. (c)

from final value theorem,

lim I(t) = lim sI (s)

= lim s x 2/s (1 + s)

= 2

  1. (d)

the coefficients cn and c-n in the fourier series are complex conjugate.

C-N = CN*

|C-N |= |CN|

<C-N = – <CN

C3 = 3 + J5

C-3 = C3* = 3 – J5

  1. (c)

transfer function h(s) = FT of output/FT if input

= 4e-j20 x(s)/x(s) = 4e-j2o

h(s) = 4e-j4f

  1. (b)

y(z) = h(z) x(z)

given,   x(z) = z4 + z2 – 2z + 2 – 3z-4

h(n) = [n – 3]

h(z) = z-3

y(z) = z-3 [z4 + z2 – 2z + 2 – 3z-4]

= z + z-1 + 2z-2 + 2z-3 – 3z-7

taking inverse z-transform,

by inspection,

y[n] = [n + 1] + [n – 1] – 2[n – 2] + 2 [n – 3] – 3 [n – 7]

putting n = 4

y[4] = 0

  1. (c)

from shifting property of impulse function,

(t)* (t – to) = (t – t0)

hence, x(t + 5)* (t – 7) = x(t + 5 – 7)

= x(t – 2)

  1. (b)

signal x(t) = 2 cos t + 7 cos t

period of 2 cos t = t1 = 2/0

= 2 = 2

period of 7 cos t = t2 = 2/0

= 2/1 = 2

for signal x(t) to be periodic t1/t2 not be rational.

t1/t2 = 2/2 = 1 = irrational

periodic waveform is sufficient condition for existence of fourier series. hence, x(t) = 2 cos t + 7 cos t is not a periodic signal hence, fourier series expansion is not possible.

  1. (c)

from duality property of fourier transform.

x(t) – x(f)

x(t) – x(-f)

f{1/1 + j2t} = ef u(-f)

  1. (a)

let   VI = x(t) cos t is applied in linear phase channel.

where, Tg = group delay

Tp = phase delay

in liear phase channel, transmission is distrtionless.

hence,   TP = TG = constant

  1. (c)

given that sampling interval ts = 100 us

sampling frequency fs = 1/ts = 10 kHz

bandwidth of LPF = fc = 5 kHz

fs = 2 B

hence, sampled output not overlap.

filter output = y(t)/ts

= 5 x 10-6 x x(t)/100 x 10-6

given,   x(t) = 10 cos (8 x 103) t

output = 5 x 10-6 x 10 cos (8 x 103) t/100 x 10-6

= 5 x 10-1 cos (8 x 103) t

  1. (b)

transfer function h(s) = 1/s2(s – 2)

lmpulse response h(t) = l-1 [h(s)]

= l-1 [1/s2(s – 2)]

we know that

l-1 [f1 (t). f2(t)] = l-1 [f1(t)* l-1[f2(t)]

hence, h(t) = l-1 [1/s2]* l-1|1/s – 2]

= t u(t)* e2t u(t)

= (t* e2t) u(t)

  1. (a)

x[n] = u[n]

x[z] = x[k] z-k

u[k] z-k = z-k

x[z] = 1/1 – z-1

for x[z] to converge,

|z-1|< 1

|z|> 1

  1. (a)

from sampling property of unit impulse function,

(t)  (t) dt = (0)

(t) cos (3t/2) dt = cos (0) = 1

  1. (d)

As band limited signal is sampled at nyquist rate. hence, replicated spectra do not overlap and the original sepctrum can be generated by passing 0s (t) through ideal low-pass filter that has cut-off frequency.

(bandwidth fc  = fs/2 ; where fs is sampling frequency.

  1. (a)

y(s) = x(s) h(s)

given,  x(t) = e3t,  h(t) = e2t

x(s) = 1/s – 3,  h(s) = 1/s – 2

y(s) = x(s)  h(s) = 1/(s – 3) x 1/(s – 2)

= 1/s – 3 – 1/s – 2

hence, y(t) = e3t –– e2t

  1. (b)

given, l[f(t)] = s + 2/s2 + 1 = f(s)

l[g(t)] = s2 + 1/(s + 3)(s + 2) = g(s)

h(t) = f g(t-) d

h(t) = f(t)* g(t)

taking laplace transform,

l[h(t) = l[f(t)* g(t)]

= l[f(t)]. l[g(t)]

= f(s). g(s)

= s + 2/s2 + 1 x s2 + 1/(s + 3) (s + 2) = 1/s + 3

  1. (b)

the signal x(t) = e-3t2 is an example of gaussian distribution. the fourier transform of gaussian distribution is also caussian distribution.

x(f) = A e-bf2

  1. (b)

let x(t) = a x1 (t) + b x2 (t)

then, a y1 (t) + b y2 (t) = at x1 (t) + bt x2(t)

= t x(t)

then system is linear.

if input is delayed by t0, i.e., x(t – t0)

then output is

y1 (t) = t x (t – to)

but output is delayed by to.

y2(t) = (t – to x (t – to)

y1 (t) = y2(t)

hence, system is not time-invariant, i.e., time varying.

  1. (b)

x[n] = xdft [k]

a typial DFT term XDFT[K] has the form Aejo

let = 2kn /n, the IDFT gives x[n] with terms of the form

1/n Aejo . ej = 1/n Aej(0+ 0)

if we conjugate XDFT [k], typical term becomes Ae-jo. its IDFT gives terms of the form Ae-joe-jo = Ae-j(0 + 0)

this DFT result corresponds to the signal Nx *[n].

  1. (c)

the DFT of a real signal shows conjugate symmetry as

XDFT[K] = XDFT [N – K]

here, N = 6

a = XDFT [1] = XDFT [6 – 1] = XDFT [5]

a = – j

B = XDFT [4] = XDFT[6 – 4] = XDFT[2]

B = 2 – J

  1. (d)

let x[n] = (1,2,1,2} is discrete fourier transform of x[n] is given as

XDFT [K] = X[n] e-j2nk/n

k = 0, 1,……….N – 1 and

n = 4

so, XDFT [K] = x[n] e-j2nk/2

x[n] e-jkn/2

k = 0, XDFT [0] = x[n] eo

x[0] + x[1] + x[2] + x[3]

= 1 + 2 + 1 + 2 = 6

k = 1, XDFT [1] = x[n] e-jn/2

= x1 [0] e0 + x[1] e-j/2 + x[2] e-j + x[3] ej3/2

= 1 – 2j – 1 + 2j = 0

k = 2, XDFT [2] = x[n] e-jn

= x[0] e0 + x[1] e-j + x[2] e-j2 + x[3] e-j3

= 1 – 2 + 1 – 2 = 2

k = 3, XDFT [3] = x[n] e-jn3/2

= x[10] e0 + x [1] e-j3/2 + x[2] e-j3 + x[3] e-j9/2

= 1 – 2j – 1 + 2j = 0

so,   XDFT [K] = {6,0, -2, 0}

  1. (a)

for a band-pass signal, minimum sampling frequency is given as

fs = 2fH/N

where,   N = int (fH/B)

Here,       fl = 4 kHz, fh = 6 kHz

bandwidth   B = FH – FL = 6 – 4 = 2 kHz

N = int (6/2) = 3

fs = 2 x 6/3 = 4 kHz

  1. (c)

x1 (t) * x2(t) = x1 (f) x2 (f) (product of spectrum)

so,    x1(t)* x2(t) extends only to 2 kHz

nyquist rate = 2 x 2 kHz = 4 kHz

  1. (a)

Because present output y(n) depends upon past output y(n – 1) and future output y(n + 1).

  1. (a)

for causal system output must depend upon present and past, not on future.

  1. (b)

The periods of individual components are

01 = 2/3 2/T1 = 2/3   t1 = 3

02 = 1/4  2/t2 = 1/4  t2 = 8

03 = 1/3  2/t3 = 1/3   t2 = 6

04 = 1/2  2/t4 = 1/2  t4 = 4

Common period is LCM of the individual periods

T = LCM (3, 8, 6, 4)

= 24